Dependence of power and efficiency of the current source on the load. Electrical circuits Questions and tasks for self-control


Let's consider the energy relationships in a closed DC circuit. In Fig. 106, a closed DC circuit was presented, powered by an element e. d.s. Ш and with internal resistance, we denote the external resistance of the circuit by R. The total power released in the circuit will be the sum of the powers released in the external and internal parts of the circuit:

W = l1R-rriR№ = ]i(R-:-Rll),

or, since by formula (For) § 164 I (R-(- R0) - £, then

Thus, the total power released in the circuit is expressed by the product of the current strength and e. d.s. element. This power is released due to any third-party energy sources; such energy sources can be, for example, chemical reactions occurring in the element.

Consequently, in a direct current circuit, external forces develop positive power 1Sh.

6 S. Frisch to A. Tiyoreva

is closed by an external resistance R-, we determine the dependence on R of the following "values: the total Power W released in the circuit, the power Wa released in the external part of the circuit, and the efficiency % which is numerically equal to the ratio of the power released in the external part of the circuit , to all power.

The current strength I in value is expressed according to Ohm’s law by the relation:

It reaches its greatest value at R = 0; in this case the current is called short circuit current, its strength is equal to:

As the external resistance increases, the current decreases, tending asymptotically to zero with an infinite increase in external resistance (see Fig. 108).

The total power released in the circuit will be:

It reaches its greatest value at short circuit current (R = 0):

Rice. 108. Dependence of current strength.„, Ш3

from external resistance. Wmax-^r

As R increases, the power decreases, tending asymptotically to zero as R increases indefinitely.

The power released in the external part of the circuit is equal to:

At short circuit current R = 0, whence the power released in the external part of the circuit is equal to zero. Wa reaches its greatest value at R = R(I, i.e. when the external resistance is equal to the internal one. In this case

i.e. equal to a quarter of the power during a short circuit.

To make sure that the maximum power Wa is obtained at R=Rt>, let’s take the derivative of Wa with respect to external resistance:

1- -(R*-R*) dR (R + Ro)4

According to the maximum condition, the first derivative must be equal to zero:

Where does R = Ra.

We can verify that under this condition we obtain a maximum and not a minimum for Wa by determining the sign of the second derivative.

With an infinite increase in external resistance, the power released in the external circuit tends to zero.

We determine the efficiency by the ratio of the power Wa released in the external part of the circuit to the total power W:

For R = 0 we have -rj = 0; with increasing R, efficiency t) increases, tending to the value i] =;l with an unlimited increase in R, however, the power released in the external circuit tends to zero, therefore the condition for maximum efficiency with not interesting from a practical point of view.

In Fig. 109 curve / gives the dependence of the power Wa released in the external part of the circuit on the resistance of the external part of the circuit R-, curve 2 "gives the dependence on R of the total power W; finally, curve 3 gives the variation of the efficiency sch from the same external resistance R. As can be seen, "] increases with increasing R.

The most interesting, from a practical point of view, power Wa, released in the external part of the circuit, first increases, and then, having reached a maximum at R = R(), begins to decrease.

At R = R0, when Wa has a maximum, =

Dependence of power and efficiency of the current source on the load

Devices and accessories: laboratory panel, two batteries, milliammeter, voltmeter, variable resistors.

Introduction. The most widely used sources of direct current are galvanic cells, batteries, and rectifiers. Let's connect to the current source the part that needs its electrical energy (light bulb, radio, microcalculator, etc.). This part of the electrical circuit is generally called the load. The load has some electrical resistance R and consumes current from the source I(Fig. 1).

The load forms the external part of the electrical circuit. But there is also an internal part of the circuit - this is actually the current source itself, it has electrical resistance r, the same current flows in it I. The boundary between the internal and external sections of the circuit are the “+” and “–” terminals of the current source, to which the consumer is connected

In Figure 1, the current source is covered by a dashed outline.

Current source with electromotive force E creates a current in a closed circuit, the strength of which is determined Ohm's law:

When current flows through resistances R And r thermal energy is released in them, determined by law Joule-Lenz. Power in the external part of the circuit R e – external power

This power is useful.

Power on the inside R i – internal power. It is not available for use and is therefore losses source power

Full current source power R is the sum of these two terms,

As can be seen from definitions (2,3,4), each of the powers depends on both the flowing current and the resistance of the corresponding part of the circuit. Let's consider this dependence separately.

Power dependenceP e , P i , P from the load current.

Taking into account Ohm's law (1), the total power can be written as follows:

Thus, The total power of the source is directly proportional current consumption.

Power released at the load ( external), There is

It is equal to zero in two cases:

1) I = 0 and 2) E – Ir = 0. (7)

The first condition is valid for an open circuit when R , the second corresponds to the so-called short circuit source when the external circuit resistance R = 0 . In this case, the current in the circuit (see formula (1)) reaches its greatest value – short circuit current.

At this current full power becomes greatest

R nb = EI short circuit =E 2 / r. (9)

However, she all stands out inside the source.

Let us find out under what conditions external power becomes maximum. Power dependence P e from the current is (see formula (6)) parabolic:

.

The position of the maximum of the function is determined from the condition:

dP e /dI = 0, dP e /dI = E – 2Ir.

The useful power reaches its maximum value at current

which is half the short circuit current (8), (see Fig. 2):

The external power at this current is

(12)

those. the maximum external power is one-fourth of the maximum total power of the source.

Power released by internal resistance during current I max is defined as follows:

, (13)

those. is also one quarter of the maximum total power of the current source. Note that at current I max

P e = P i . (14)

When the current in the circuit tends to its greatest value I short circuit , internal power

those. equal to the highest power of the source (9). This means that all the power of the source is allocated to its internal resistance, which, of course, is harmful from the point of view of safety of the current source.

Characteristic points of the dependence graph P e = P e (I) shown in Fig. 2.

Efficiency the operation of the current source is estimated efficiency. Efficiency is the ratio of useful power to the total power of the source:

= P e / P.

Using formula (6), the expression for efficiency can be written as follows:

. (15)

From formula (1) it is clear that E – Ir = IR there is tension U on external resistance. Therefore, efficiency

= U/ E . (16)

From expression (15) it also follows that

= (17)

those. The efficiency of the source depends on the current in the circuit and tends to the highest value, equal to unity, at current I 0 (Fig.3) . As the current increases, the efficiency decreases linearly and goes to zero during a short circuit, when the current in the circuit becomes greatest. I short circuit = E/ r .

From the parabolic nature of the dependence of external power on current (6) it follows that the same power on the load P e can be obtained at two different current values ​​in the circuit. From formula (17) and from the graph (Fig. 3) it is clear that in order to obtain greater efficiency from the source, it is preferable to operate at lower load currents, where this coefficient is higher.

2.Power dependenceP e , P i , P from load resistance.

Let's consider addiction complete, useful and internal power from external resistanceR in the source circuit with EMF E and internal resistance r.

Full the power developed by the source can be written as follows if we substitute the expression for current (1) into formula (5):

So the total power depends on the load resistance R. It is greatest during a short circuit, when the load resistance goes to zero (9). With increasing load resistance R The total power decreases, tending to zero at R .

Stands out at the external resistance

(19)

External power R e is part of the total power R and its value depends on the resistance ratio R/(R+ r) . During a short circuit, the external power is zero. As resistance increases R it first increases. At R  r external power tends to full in magnitude. But the useful power itself becomes small, since the total power decreases (see formula 18). At R  external power tends to zero as does total power.

What should be the load resistance to receive from this source maximum external (useful) power (19)?

Let's find the maximum of this function from the condition:

Solving this equation, we get R max = r.

Thus, Maximum power is released in the external circuit if its resistance is equal to the internal resistance of the current source. Under this condition, the current in the circuit is equal to E/2 r, those. half the short circuit current (8). Maximum useful power at this resistance

(21)

which coincides with what was obtained above (12).

Power released at the internal resistance of the source

(22)

At R P i P, and when R=0 reaches its greatest value P i nb = P nb = E 2 / r. At R= r internal power is half full, P i = P/2 . At R r it decreases almost in the same way as the full one (18).

The dependence of efficiency on the resistance of the external part of the circuit is expressed as follows:

 = (23)

From the resulting formula it follows that the efficiency tends to zero as the load resistance approaches zero, and the efficiency tends to the highest value equal to unity as the load resistance increases to R r. But the useful power decreases almost as much as 1/ R (see formula 19).

Power R e reaches its maximum value at R max = r, the efficiency is equal, according to formula (23), = r/(r+ r) = 1/2. Thus, the condition for obtaining maximum useful power does not coincide with the condition for obtaining the greatest efficiency.

The most important result of the consideration is the optimal matching of the source parameters with the nature of the load. Three areas can be distinguished here: 1) R r, 2)R r, 3) R r. First the case occurs where low power is required from the source for a long time, for example, in electronic watches, microcalculators. The size of such sources is small, the supply of electrical energy in them is small, it must be spent economically, so they must operate with high efficiency.

Second case - a short circuit in the load, in which all the power of the source is released in it and the wires connecting the source to the load. This leads to excessive heating and is a fairly common cause of fires and fires. Therefore, a short circuit of high power current sources (dynamos, batteries, rectifiers) is extremely dangerous.

IN third case, they want to get maximum power from the source at least for a short time, for example, when starting a car engine using an electric starter, the efficiency value is not so important. The starter turns on for a short time. Long-term operation of the source in this mode is practically unacceptable, as it leads to rapid discharge of the car battery, its overheating and other troubles.

To ensure the operation of chemical current sources in the required mode, they are connected to each other in a certain way into so-called batteries. The elements in the battery can be connected in series, parallel or in a mixed circuit. This or that connection scheme is determined by the load resistance and the amount of current consumed.

The most important operational requirement for power plants is their high efficiency. From formula (23) it is clear that the efficiency tends to unity if the internal resistance of the current source is small compared to the load resistance

In parallel, you can connect elements that have the same EMF. If connected n identical elements, then from such a battery you can get current

Here r 1 – resistance of one element, E 1 – EMF of one element.

Such a connection is advantageous to use with low-resistance loads, i.e. at R r. Since the total internal resistance of the battery when connected in parallel decreases by n times compared to the resistance of one element, then it can be made close to the load resistance. Thanks to this, the efficiency of the source increases. Increases in n times and the energy capacity of the battery elements.

 r, then it is more profitable to connect the elements in a battery in series. In this case, the emf of the battery will be n times greater than the EMF of one element and the required current can be obtained from the source

Purpose this laboratory work is experimental verification The theoretical results obtained above on the dependence of the total, internal and external (net) power and the efficiency of the source on both the power of the consumed current and the load resistance.

Description of installation. To study the operating characteristics of the current source, an electrical circuit is used, the diagram of which is shown in Fig. 4. Two NKN-45 alkaline batteries are used as a current source, which are connected sequentially into one battery through a resistor r , modeling the internal resistance of the source.

Its inclusion artificially increases the internal resistance of the batteries, which 1) protects them from overload when switching to short circuit mode and 2) makes it possible to change the internal resistance of the source at the request of the experimenter. As a load (external circuit resistance) p
two variable resistors are used R 1 And R 2 . (one coarse adjustment, the other fine), which provides smooth current regulation over a wide range.

All instruments are mounted on a laboratory panel. The resistors are fixed under the panel; their control knobs and terminals are located at the top, near which there are corresponding inscriptions.

Measurements. 1.Install the switch P to neutral position, switch VC open. Turn the resistor knobs counterclockwise until they stop (this corresponds to the highest load resistance).

    Assemble the electrical circuit according to the diagram (Fig. 4), do not joining for now current sources.

    After checking the assembled circuit by a teacher or laboratory assistant, connect the batteries E 1 And E 2 , observing polarity.

    Set the short circuit current. To do this, set the switch P to position 2 (external resistance is zero) and using a resistor r set the milliammeter needle to the limit (rightmost) division of the instrument scale - 75 or 150 mA. Thanks to the resistor r in the laboratory setup there is possibility to regulate internal resistance of the current source. In fact, internal resistance is a constant value for this type of source and cannot be changed.

    Set the switch P to position 1 , thereby turning on the external resistance (load) R= R 1 + R 2 into the source circuit.

    Changing the current in the circuit through 5...10 mA from the highest to the lowest value using resistors R 1 And R 2 , record the milliammeter and voltmeter readings (load voltage U) into the table.

    Set the switch P to neutral position. In this case, only a voltmeter is connected to the current source, which has a fairly large resistance compared to the internal resistance of the source, so the voltmeter reading will be slightly less than the emf of the source. Since you have no other way to determine its exact value, it remains to take the voltmeter reading as E. (See lab #311 for more information on this.)

pp

mA

P e ,

P i ,

R,

Processing the results. 1. For each current value, calculate:

    total power according to formula (5),

    external (useful) power according to the formula,

    internal power from the ratio

    resistance of the external section of the circuit from Ohm's law R= U/ I,

    Efficiency of the current source according to formula (16).

    Build dependency graphs:

    total, useful and internal power from current I (on one tablet),

    total, useful and internal power from resistance R(also on one tablet); it is more reasonable to construct only part of the graph corresponding to its low-resistivity part, and discard 4-5 experimental points out of 15 in the high-resistivity region,

    Source efficiency versus current consumption I,

    Efficiency versus load resistance R.

    From graphs P e from I And P e from R determine the maximum net power in the external circuit P e max.

    From the graph P e from R determine the internal resistance of the current source r.

    From graphs P e from I And P e from R find the efficiency of the current source at I max and at R max .

Control questions

1.Draw a diagram of the electrical circuit used in the work.

2.What is a current source? What is the load? What is the internal section of the chain? Where does the outer section of the chain begin and end? Why is a variable resistor installed? r ?

3.What is called external, useful, internal, total power? How much power is lost?

4. Why is it proposed to calculate the useful power in this work using the formula P e = IU, and not according to formula (2)? Justify these recommendations.

5. Compare the experimental results you obtained with the calculated ones given in the methodological manual, both when studying the dependence of power on current and on load resistance.

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    • OHM'S LAW FOR A COMPLETE CIRCUIT:

    I is the current strength in the circuit; E is the electromotive force of the current source connected to the circuit; R - external circuit resistance; r is the internal resistance of the current source.

    POWER DELIVERED IN THE EXTERNAL CIRCUIT

    . (2)

    From formula (2) it is clear that in the event of a short circuit ( R®0) and at R® this power is zero. For all other final values R power R 1 > 0. Therefore, the function R 1 has a maximum. Meaning R 0, corresponding to the maximum power, can be obtained by differentiating P 1 with respect to R and equating the first derivative to zero:

    . (3)

    From formula (3), taking into account the fact that R and r are always positive, and E? 0, after simple algebraic transformations we get:

    Hence, the power released in the external circuit reaches its greatest value when the resistance of the external circuit is equal to the internal resistance of the current source.

    In this case, the current strength in the circuit (5)

    equal to half the short circuit current. In this case, the power released in the external circuit reaches its maximum value equal to

    When the source is closed to an external resistance, then current flows inside the source and at the same time a certain amount of heat is released at the internal resistance of the source. The power expended to release this heat is equal to

    Consequently, the total power released in the entire circuit is determined by the formula

    = I 2(R+r) = I.E. (8)

    EFFICIENCY

    EFFICIENCY current source is equal . (9)

    From formula (8) it follows that

    those. R 1 changes with the change in current in the circuit according to a parabolic law and takes zero values ​​at I = 0 and at . The first value corresponds to an open circuit (R>> r), the second to a short circuit (R<< r). Зависимость к.п.д. от силы тока в цепи с учётом формул (8), (9), (10) примет вид

    Thus, the efficiency reaches its highest value h =1 in the case of an open circuit (I = 0), and then decreases according to a linear law, becoming zero in the case of a short circuit.

    Dependence of powers P 1, P full = EI and efficiency. current source and the current strength in the circuit are shown in Fig. 1.

    Fig.1. I 0 E/r

    From the graphs it is clear that to obtain both useful power and efficiency. impossible. When the power released in the external section of the circuit P 1 reaches its greatest value, efficiency. at this moment it is 50%.

    METHOD AND PROCEDURE OF MEASUREMENTS


    Assemble the circuit shown in Fig. on the screen. 2. To do this, first click the left mouse button above the emf button. at the bottom of the screen. Move the mouse marker to the working part of the screen where the dots are located. Click the left mouse button in the working part of the screen where the emf source will be located.

    Next, place a resistor in series with the source, representing its internal resistance (by first pressing the button at the bottom of the screen) and an ammeter (the button is in the same place). Then arrange the load resistors and voltmeter in the same way, measuring the voltage across the load.

    Connect the connecting wires. To do this, click the wire button at the bottom of the screen, and then move the mouse marker to the working area of ​​the circuit. Click with the left mouse button in the areas of the working area of ​​the screen where the connecting wires should be located.

    4. Set parameter values ​​for each element. To do this, left-click on the arrow button. Then click on this element. Move the mouse marker to the slider of the regulator that appears, click on the left mouse button and, holding it down, change the parameter value and set the numerical value indicated in Table 1 for your option.

    Table 1. Initial parameters of the electrical circuit

    option

    5. Set the external circuit resistance to 2 Ohms, press the “Count” button and write down the readings of electrical measuring instruments in the corresponding lines of Table 2.

    6. Use the regulator slider to consistently increase the resistance of the external circuit by 0.5 Ohms from 2 Ohms to 20 Ohms and, pressing the “Count” button, record the readings of electrical measuring instruments in Table 2.

    7. Calculate using formulas (2), (7), (8), (9) P 1, P 2, P total and h for each pair of voltmeter and ammeter readings and write the calculated values ​​in Table 2.

    8. Construct on one sheet of graph paper graphs of the dependence P 1 = f (R), P 2 = f (R), P total = f (R), h = f (R) and U = f (R).

    9. Calculate the measurement errors and draw conclusions based on the results of the experiments.

    Table 2. Results of measurements and calculations

    P full, VT

    Questions and tasks for self-control

    1. Write the Joule-Lenz law in integral and differential forms.
    2. What is short circuit current?
    3. What is gross power?
    4. How is efficiency calculated? current source?
    5. Prove that the greatest useful power is released when the external and internal resistances of the circuit are equal.
    6. Is it true that the power released in the internal part of the circuit is constant for a given source?
    7. A voltmeter was connected to the flashlight battery terminals, which showed 3.5 V.
    8. Then the voltmeter was disconnected and a lamp was connected in its place, on the base of which it was written: P = 30 W, U = 3.5 V. The lamp did not burn.
    9. Explain the phenomenon.
    10. When the battery is alternately shorted to resistances R1 and R2, an equal amount of heat is released in them at the same time. Determine the internal resistance of the battery.

    LABORATORY WORK No. 3.7.

    STUDY OF USEFUL POWER AND EFFICIENCY OF CURRENT SOURCES

    Last name I.O. _____________ Group ______ Date ______

    Introduction

    The purpose of this work is to experimentally test the theoretical conclusions about the dependence of the useful power and efficiency of the current source on the load resistance.

    An electrical circuit consists of a current source, supply wires and a load or current consumer. Each of these circuit elements has resistance.

    The resistance of the lead wires is usually very small, so it can be neglected. In each section of the circuit, the energy of the current source will be consumed. The question of the appropriate use of electrical energy is of very important practical importance.

    The total power P released in the circuit will be the sum of the powers released in the external and internal parts of the circuit: P = I 2 R + I 2 r = I 2 (R + r). Because I(R + r) = ε, That Р =I·ε,

    where R is external resistance; r – internal resistance; ε – EMF of the current source.

    Thus, the total power released in the circuit is expressed by the product of the current and the emf of the element. This power is released due to any third-party energy sources; such energy sources can be, for example, chemical processes occurring in the element.

    Let's consider how the power released in the circuit depends on the external resistance R to which the element is closed. Let us assume that an element of a given EMF and a given internal resistance r is closed by an external resistance R; Let us determine the dependence on R of the total power P allocated in the circuit, the power Ra allocated in the external part of the circuit and the efficiency.

    The current strength I in the circuit is expressed according to Ohm's law by the relation

    The total power released in the circuit will be equal to

    As R increases, the power decreases, tending asymptotically to zero as R increases indefinitely.

    The power released in the external part of the circuit is equal to

    From this it can be seen that the useful power P a is equal to zero in two cases - at R = 0 and R = ∞.

    Exploring the function R a = f(R) to the extremum, we find that P a reaches a maximum at R = r, then

    To make sure that the maximum power P a is obtained at R = r, let’s take the derivative of P a with respect to external resistance

    Where

    According to the maximum condition, the first derivative must be equal to zero

    r 2 = R 2

    R = r

    You can make sure that under this condition we will obtain a maximum, and not a minimum, for P a by determining the sign of the second derivative.

    The efficiency factor (efficiency) η of an EMF source is the ratio of the power P a released in the external circuit to the total power P developed by the EMF source.

    In essence, the efficiency of an EMF source indicates what proportion of the work of external forces is converted into electrical energy and transferred to the external circuit.

    Expressing power in terms of current I, the potential difference in the external circuit U and the magnitude of the electromotive force ε, we obtain

    That is, the efficiency of the EMF source is equal to the ratio of the voltage in the external circuit to the EMF. Under the conditions of applicability of Ohm's law, one can further replace U = IR; ε = I(R + r), Then

    Consequently, in the case when all the energy is spent on Lenz-Joule heat, the efficiency of the EMF source is equal to the ratio of the external resistance to the total resistance of the circuit.

    At R = 0 we have η = 0. With increasing R, the efficiency increases and tends to the value η = 1 with an unlimited increase in R, but at the same time the power released in the external circuit tends to zero. Thus, the requirements for simultaneously obtaining maximum useful power with maximum efficiency are impossible to meet.

    When P a reaches its maximum, then η = 50%. When the efficiency η is close to unity, the useful power is small compared to the maximum power that a given source could develop. Therefore, to increase efficiency, it is necessary, if possible, to reduce the internal resistance of the EMF source, for example, a battery or a dynamo.

    In the case of R = 0 (short circuit) P a = 0 and all the power is released inside the source. This can lead to overheating of the internal parts of the source and its failure. For this reason, short circuits of sources (dynamos, batteries) are not allowed!

    In Fig. 1, curve 1 gives the dependence of the power P a released in the external circuit on the resistance of the external part of the circuit R; curve 2 gives the dependence of the total power P on R; curve 3 – variation of efficiency η from the same external resistance.

    Work order

    1. Check out the diagram at the stand.

    2. Using the magazine, set the resistance R = 100 Ohm.

    3. Close key K.

    4. Measure the current in the circuit sequentially for nine different resistances on a resistance magazine, starting from 100 Ohms and above. Enter the results of current measurements into the table, expressing them in amperes.

    5. Turn off key K.

    6. Calculate for each resistance P, P a (in watts) and η.

    7. Construct graphs of P, P a and η from R.

    Control questions

    1. What is the efficiency of an EMF source?

    2. Derive the formula for the efficiency of the EMF source.

    3. What is the useful power of an EMF source?

    4. Derive the formula for the useful power of the EMF source.

    5. What is the maximum power released in the external circuit (Pa)max?

    6. At what value of R is the total power P released in the circuit maximum?

    7. What is the efficiency of the EMF source at (Pa)max?

    8. Carry out a study of the function (Pa) = f(R) to the extreme.

    9. Draw a graph of the dependence of P, Ra and η on the external resistance R.

    10. What is source emf?

    11. Why should external forces be of non-electrical origin?

    12. Why is short circuiting unacceptable for voltage sources?

    No.

    R,Ohm

    I·10 -3 ,A

    , W

    , W

    1

    0

    2

    100

    3

    200

    4

    300

    5

    400

    6

    500

    7

    600

    8

    700

    9

    800

    10

    900

    r = 300Ohm

    1. What is the time it takes for a current of 5 A to pass through a conductor if, at a voltage at its ends of 120 V, an amount of heat equal to 540 kJ is released in the conductor? (Give your answer in seconds.)

    2. In an electric heater with a constant coil resistance, through which a direct current flows, for a time t amount of heat released Q. If the current strength and time t doubled, then how many times will the amount of heat released in the heater increase?

    3. Resistor 1 with an electrical resistance of 3 Ohms and resistor 2 with an electrical resistance of 6 Ohms are connected in series in a DC circuit. What is the ratio of the amount of heat released by resistor 1 to the amount of heat released by resistor 2 in the same time?

    4. The figure shows a graph of the dependence of the current in an incandescent lamp on the voltage at its terminals. What is the current power in the lamp at a voltage of 30 V? (Give your answer in watts.)

    5.

    The student assembled the electrical circuit shown in the figure. What energy will be released in the external part of the circuit when current flows for 10 minutes? (Express your answer in kJ. The necessary data is indicated in the diagram. Consider the ammeter ideal.)

    6. A capacitor with a capacity of 1 μF is connected to a current source with an emf of 2 V. How much work was done by the source when charging the capacitor? (Give your answer in µJ.)

    7. A capacitor with a capacity of 1 μF is connected to a current source with an emf of 2 V. How much heat will be released in the circuit during the charging of the capacitor? (Give your answer in µJ.) Neglect radiation effects.

    8. A capacitor with a capacity of 1 μF is connected to an ideal current source with an emf of 3 V once through a resistor and the second time - through a resistor How many times more is the heat generated by the resistor in the second case compared to the first? Neglect radiation.

    9. To a current source with EMF 4 V and internal resistance connected the load resistor. What should it be equal for the source efficiency to be 50%? (Give your answer in ohms.)

    10. In the electrical circuit, the diagram of which is shown in the figure, the measuring instruments are ideal, the voltmeter shows a voltage value of 8 V, and the ammeter shows a current value of 2 A. How much heat will be released in the resistor in 1 second? (Give your answer in joules.)

    11. The room is illuminated by four identical bulbs connected in parallel. Electricity consumption per hour is Q. What should be the number of parallel-connected light bulbs so that the electricity consumption per hour is equal to 2 Q?

    12. An electric kettle with a power of 2.2 kW is designed to be connected to an electrical network with a voltage of 220 V. Determine the current strength in the heating element of the kettle when it is operating in such a network. Give your answer in amperes.

    13. On the body of the electric roaster there is an inscription: “220 V, 660 W.” Find the current consumed by the roaster. (Give your answer in amperes.)

    14. On the base of an electric incandescent lamp it is written: “220 V, 60 W.” Two such lamps are connected in parallel and connected to a voltage of 127 V. What power will be released in these two lamps with this connection method? (Give your answer in watts, rounded to the nearest whole number.) When solving the problem, assume that the resistance of the lamp does not depend on the voltage applied to it.

    15. On the base of an electric incandescent lamp it is written: “220 V, 100 W.” Three such lamps are connected in parallel and connected to a voltage of 127 V. What power will be released in these three lamps with this connection method? (Give your answer in watts, rounded to the nearest whole number.) When solving the problem, assume that the resistance of the lamp does not depend on the voltage applied to it.

    16. There are two round conductors in the school laboratory. The resistivity of the first conductor is 2 times greater than the resistivity of the second conductor. The length of the first conductor is 2 times the length of the second. When these conductors are connected to the same constant voltage sources at the same time intervals, the amount of heat released in the second conductor is 4 times greater than in the first. What is the ratio of the radius of the second conductor to the radius of the first conductor?

    17. There are two round conductors in the school laboratory. The resistivity of the first conductor is 2 times greater than the resistivity of the second conductor. The length of the first conductor is 2 times the length of the second. When these conductors are connected to the same constant voltage sources at the same time intervals, the amount of heat released in the second conductor is 4 times less than in the first. What is the ratio of the radius of the first conductor to the radius of the second conductor?

    18. R 1, included in the electrical circuit, the diagram of which is shown in the figure? (Answer in watts.) R 1 = 3 Ohm, R 2 = 2 Ohm, R

    19. How much power is released in the resistor R 2, included in the electrical circuit, the diagram of which is shown in the figure? (Answer in watts.) R 1 = 3 Ohm, R 2 = 2 Ohm, R 3 = 1 Ohm, source emf 5 V, internal resistance of the source is negligible.

    20. R= 16 Ohm, and the voltage between the points A And B equal to 8 V? Give your answer in watts.

    21. What power is released in the section of the circuit, the diagram of which is shown in the figure, if R= 27 Ohm, and the voltage between the points A And B equal to 9 V? Give your answer in watts.

    22. I= 6 A. What is the current strength shown by the ammeter? (Give your answer in amperes.) Neglect the resistance of the ammeter.

    23. A resistor with resistance is connected to a current source with EMF and internal resistance. If you connect this resistor to a current source with EMF and internal resistance then how many times will the power released in this resistor increase?

    24.

    I U on the lamp. Such a lamp is connected to a constant voltage source of 2 V. How much work will the electric current do in the filament of the lamp in 5 seconds? Express your answer in J.

    25.

    The graph shows the experimentally obtained dependence of the current strength I, flowing through an incandescent lamp, from voltage U on the lamp. Such a lamp is connected to a constant voltage source of 4 V. How much work will the electric current do in the filament of the lamp in 10 seconds? Express your answer in J.

    26. A direct current flows through a section of the circuit (see figure) I= 4 A. What current will be shown by an ideal ammeter connected to this circuit if the resistance of each resistor r= 1 Ohm? Express your answer in amperes.

    27. A point positive charge of 2 µC is placed between two extended plates, uniformly charged with opposite charges. The modulus of the electric field strength created by a positively charged plate is 10 3 kV/m, and the field created by a negatively charged plate is 2 times greater. Determine the magnitude of the electric force that will act on the indicated point charge.

    28. A point positive charge of 2 µC is placed between two extended plates, uniformly charged with positive charges. The modulus of the electric field strength created by one plate is 10 3 kV/m, and the field created by the second plate is 2 times greater. Determine the magnitude of the electric force that will act on the indicated point charge. Give your answer in newtons.

    29.

    WITH, resistor resistance R and key K. The capacitor is charged to voltage U= 20 V. The charge on the capacitor plates is q= 10 –6 Cl. How much heat will be released in the resistor after switch K is closed? Express your answer in mJ.

    30.

    The figure shows a diagram of an electrical circuit consisting of a capacitor with a capacity WITH, resistor resistance R and key K. Capacitance of the capacitor C= 1 µF and it is charged to voltage U= 10 V. How much heat will be released in the resistor after closing key K? Express your answer in mJ.

    31. The fuse of the electricity meter in a residential network with a voltage of 220 V is equipped with the inscription: “6 A”. What is the maximum total power of electrical appliances that can be connected to the network at the same time without the fuse melting? (Give your answer in watts)
    

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